Chi sqyre test

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Chi sqyre test

  1. 1. M.Prasad Naidu MSc Medical Biochemistry, Ph.D,.
  2. 2.  It is a non parametric test , not based on any assumption or distribution of any variable.  It is very useful in research and it is most commonly used when the data are given in frequencies .  It can be used with any data which can be reduced to proportion or percentages.  This test involves calculation of quantity called chi- square , which derived from Greek letter ( χ)² and pounced as ‘kye’. Cont…
  3. 3. It is an alternate test to find the significance of difference in two or more than two proportions. Chi –square test is yet another useful test which can be applied to find the significance in the same type of data with the following advantages .  cont…
  4. 4. To compare the values of two binomial samples even if they are small such as incidence of diabetes in 20 obese persons with 20 non obese persons. To compare the frequencies of two multinomial samples such as number of diabetes and non diabetes in groups weighing 40-50, 50-60, 60-70, and > 70 kg s of weight.
  5. 5.  Test of association between events in binomial or multinomial samples .  Two events often can be tested for their association such as cancer and smoking .  Treatment and outcome of disease .  Vaccination and immunity .  Nutrition and intelligence .  Cholesterol and heart disease .  Weight and diabetes, B P and heart disease.  To find they are independent of each other or dependent on each other i.e. associated .
  6. 6.  Three essentials to apply chi - square test are  A random sample  Qualitative data  Lowest frequency not less than 5  Steps :-  Assumption of Null Hypothesis (HO)  Prepare a contingency table and note down the observed frequencies or data (O)  cont…
  7. 7.  Determine the expected number (E) by multiplying CT × RT /GT (column total ,row total and grand total )  Find the difference between observed and expected frequencies in each cell (O-E)  Calculate chi- square value for each cell with  (O-E)²/E  Sum up all chi –square values to get the total chi-square value (χ)² d.f. (degrees of freedom)  = χ²= ∑(O-E)²/E and d.f. is (c-1) (r-1)
  8. 8.  Chi- square test applied in a four fold table will not give reliable result , with one degree of freedom.  If the observed value of any cell is < 5 in such cases Yates correction can be applied by subtracting ½ (χ)² = ∑ (O-E-1/2)²/E  Even with Yates correction the test may be misleading if any expected value is much below 5 in such cases Yates correction can not be applied.  cont…
  9. 9.  In tables larger than 2 × 2 Yates correction can not be applied .  The highest value of chi- square χ² obtainable by chance or worked out and given in (χ)² table at different degrees of freedom under probabilities (P) such as 0.05, 0.01, 0.001 .  If calculated value of chi-square of the sample is found to be higher than the expected value of the table at critical level of significance i.e. probability of 0.05 the H O of no difference between two proportions or the H O of independence of two characters is rejected.  If the calculated value is lower the hypothesis of no difference is accepted.
  10. 10.  Groups Died Survived Total  A (a) 10 (b) 25 35  B (c) 5 (d) 60 65  Total 15 85 100  Expected value can be computed with CT×RT/GT for (a) 15×35/100= 5.25  (b) 85×35/100= 29.75  (c) 15×65/100= 9.75  (d) 85×65/100= 55.25
  11. 11.  χ²= ∑(O-E)²/E d.f. =( c-1) (r-1)   (A) =(10-5.25)²/5.25 = (4.75)²/5 = 4.30  (B) = (25-29.75)²/29.75 =(5)²/29.75 = 0 .76  (C) = (5-9.75)²/9.75 = (5)²/9.75 = 2.31  (D) = (60-55.25)²/55.25 = (5)²/55.25 = 0.40 7.77 The calculated chi- square value is 7.77 is more than Chi- square table value with 1 d.f. At 0.01, 6.64 which significant. It can be said that there is significant difference between two groups.
  12. 12.  Chi- square can be calculated in the following way also.  (χ)² = ( ad-bc)²×N (a b × c d ×ac ×b d) = (600-125 )² ( 35× 65× 15× 85) = (475)² x100 225625 00 /2900625 (2900625) = 7.77 Here the calculated chi-square value7.77 is > 6.64 at 0.01 with 1 d.f. shows significant difference b/n

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